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Suppose $X,Y, Z$ are random variables, and without loss of generality let's assume $X=f(Y, Z)$. Is the following statement true in general?

$$ E[X|Y]=0 \iff E[X|Y,Z]=0$$

I seem to have a proof (might be flawed). But that conclusion just doesn't seem natural to me.

  • $\Leftarrow$: by the law of total expectation, $$E[X|Y]=E_Z [E[X|Y,Z]=E_Z[0]=0$$
  • $\Rightarrow$: by the law of total probability, $$E[X|Y,Z]=\int_x xp(x|y,z)dx=\int_x \frac{xp(x|y)}{p(z)}dx=0$$
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    First part is fine. The second part is not correct. However, it's easy to fix, since what you're trying to prove is not correct. – 2017-02-07
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    @Paul Can you give a counterexample? – 2017-02-07
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    Let $X=YZ$, with $Y$ be uniformly distributed on $[0,1]$ and $Z$ be uniformly distributed on $[-1,1]$ and $Y$ be independent of $Z$. – 2017-02-07
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    "without loss of generality let's assume $X=f(Y, Z)$" This is a huge loss of generality (I am not sure whether you want to solve the general case or this restricted case, but anyway). – 2017-02-07
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    As far as counterexamples go, try $(X,Y)$ independent and $$Z=X$$ – 2017-02-07

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The $E[X|Y,Z]=0 \Rightarrow E[X|Y]=0$ direction is true, as you proved.

The other direction isn't. An easy way to construct counterexamples is to use degenerate random variables (r.v.s that are constant with probability $1$).

So let $X=Y+Z$ and $Y=1$ and $Z:\left(\genfrac{}{}{0pt}{}{-3/2}{1/2}\ \genfrac{}{}{0pt}{}{-1/2}{1/2}\right)$. Thus $Y$, being constant, is indepentent of both $X$ and $Z$, and we can say $X=1+Z$ (almost surely).

Then $E[X\mid Y]=E[1+Z]=0$. However, $E[X\mid Y,Z]=E[1+Z \mid Z]=1+Z \neq 0$.

Paul's example in the comments is nicer as it doesn't rely on degenerate cases, however I wanted to note an easy way to test the truthfulness of your statements.